int_{frac{pi}{6}}^{frac{pi}{3}} frac{cos^{fra | vedclass

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Question

(C) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx$. Using the property

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(C) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx$. Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we have $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$. So,$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}}(\frac{\pi}{2}-x)}{\cos^{\frac{3}{2}}(\frac{\pi}{2}-x) + \sin^{\frac{3}{2}}(\frac{\pi}{2}-x)} \, dx$. Since $\cos(\frac{\pi}{2}-x) = \sin x$ and $\sin(\frac{\pi}{2}-x) = \cos x$,we get $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \, dx$. Adding the two expressions for $I$: $2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx$. $2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$. Therefore,$I = \frac{\pi}{12}$.

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx = $ . . . . . . .

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